Chapter 10
Comparisons Involving Means,
Experimental Design, and
Analysis of Variance
Learning Objectives
1. Be able to develop interval estimates and conduct hypothesis tests about the difference between two
population means when
1
and
2
are known.
2. Know the properties of the sampling distribution of
12
xx−
.
3. Be able to use the t distribution to conduct statistical inferences about the difference between two
population means when
1
and
2
are unknown.
4. Learn how to analyze the difference between two population means when the samples are
independent and when the samples are matched.
5. Understand the basic principles of an experimental study and how they are used in a completely
randomized design.
6. Understand how the analysis of variance (ANOVA) procedure can be used to test for the equality of
k population means using data from a completely randomized design as well as data from an
observational study.
7. Know the assumptions necessary to use the analysis of variance procedure.
8. Know how to set up an ANOVA table and interpret the entries in the table.
9. Be able to use output from Excel to solve experimental design problems.
Solutions:
1. a.
x x
1 2
−
= 13.6 – 11.6 = 2
b.
/2 .05
zz
==
1.645
22
12
c.
/2 .025
zz
==
1.96
22
(2.2) (3)
2 1.17 (.83 to 3.17)
2. a.
( )
1 2 0
2 2 2 2
12
12
(25.2 22.8) 0 2.03
(5.2) 6
40 50
x x D
z
nn
−− −−
= = =
+
+
b. p-value = 1.0000 – .9788 = .0212
3. a.
( )
1 2 0
2 2 2 2
12
12
(104 106) 0 1.53
(8.4) (7.6)
80 70
x x D
z
nn
−− −−
= = = −
+
+
b. p-value = 2(.0630) = .1260
4. a.
1
= population mean for smaller cruise ships
2
= population mean for larger cruise ships
12
xx−
= 85.36 – 81.40 = 3.96
b.
22
12
.025
12
znn
+
22
(4.55) (3.97)
1.96 1.88
37 44
+=
xx−
12
b.
22 22
12
/2
12
(35) (20)
2.576 17.08
40 30
znn
+ = + =
6.
1
= mean hotel price in Atlanta
2
H0:
12
0
−
Ha:
12
0
−
( )
1 2 0
2 2 2 2
12
12
(91.71 101.13) 0 1.81
20 25
35 40
x x D
z
nn
−− −−
= = = −
+
+
H0:
m
1–
m
2=0
Ha:
m
1–
m
2¹0
b.
12
xx−
= 79 – 71 = 8
z=x1–x2
( )
–D0
s
1
2
n1
+
s
2
2
n2
=(79 –71) –0
122
25 +122
30
=2.46
For this two-tailed test, p–value is two times the upper-tail area at z = 2.46.
(79–71) ±1.96 122
25 +122
30
8
6.37 (1.63 to 14.37)
12
12
( )
12
2 2 2 2
12
12
(76 73) 2.74
66
60 60
xx
z
nn
−−
= = =
+
+
p-value = area in upper tail at z = 2.74
p-value = 1.0000 – .9969 = .0031
Since .0031
α = .05, we reject the null hypothesis. The difference is significant. We can conclude
( )
1
22
1
75.7 (76 75.7) .39
6
60
x
z
n
−−
= = =
p-value = area in upper tail at z = .39
p-value = 1.0000 – .6517 = .3483
Since .3483 > α = .05, we cannot reject the null hypothesis. The difference is not statistically
significant.
12
12
( )
12
2 2 2 2
12
12
(77 75) 1.83
66
60 60
xx
z
nn
−−
= = =
+
+
p-value = area in upper tail at z = 1.83
12
( )
12
22
1.645
66
60 60
xx
z−
==
+
22
12
66
1.645 1.80
60 60
xx− = + =
This tells us that as long as the 2008 score for a company exceeds the 2007 score by 1.80 or more the
difference will be statistically significant.
b.
22
22 22
12
12
2 2 2 2
2 2 2 2
12
1 1 2 2
2.5 4.8
20 30 45.8
1 2.5 1 4.8
11
19 20 29 30
11
ss
nn
df ss
n n n n
++
= = =
+
+
−−
d. 2.4 2.1 (.3 to 4.5)
10. a.
( )
12
2 2 2 2
12
12
0(13.6 10.1) 0 2.18
5.2 8.5
35 40
xx
tss
nn
−− −−
= = =
+
+
b.
22
22 22
12
12
2 2 2 2
2 2 2 2
12
1 1 2 2
5.2 8.5
35 40 65.7
1 5.2 1 8.5
11
34 35 39 40
11
ss
nn
df ss
n n n n
++
= = =
+
+
−−
Exact p-value corresponding to t = 2.18 is .0329
d. p-value
.05, reject H0.
11. a.
1
54 9
6
x==
2
42 7
6
x==
2
2
2
2
()
1.79
1
i
xx
sn
−
==
−
xx−
12
22
12
2.28 1.79
1.833 66
xx− +
2 2.17 (-.17 to 4.17)
12. a.
12
xx−
= 22.5 – 18.6 = 3.9
22
22 22
12
12
2 2 2 2
2 2 2 2
12
1 1 2 2
8.4 7.4
50 40 87.1
1 8.4 1 7.4
11
49 50 39 40
11
ss
nn
n n n n
++
+
+
−−
Use df = 87, t.025 = 1.988
22
8.4 7.4
3.9 1.988 50 40
+
3.9 (.6 to 7.2)
x2=Sxi
n2
=267.6 =22.3
s2=S(xi–x2)2
n2
=225.96
12 –1=4.53
b.
12
xx−
= 42.5 – 22.3 = 20.2 or $20,200
The mean annual cost to attend private colleges is $20,200 more than the mean annual cost to attend
public colleges.
s1
2
n1
+s2
2
n2
æ
è
çö
ø
÷
2
6.982
10 +4.532
12
æ
è
çö
ø
÷
2
d. p-value
.05, reject H0. We conclude that the salaries of staff nurses are lower in Tampa than in
Dallas.
15. a.
1 = population mean annual lease rate per square meter in Hong Kong
2 = population mean annual lease rate per square meter in Paris
xx−
12
t=x1–x2
( )
–0
s1
2
n1
+s2
2
n2
=(1114 –989)–0
2302
30 +1952
40
=2.40
df =
s1
2
n1
+s2
2
n2
æ
è
çö
ø
÷
2
1
n1–1
s1
2
n1
æ
è
çö
ø
÷
2
+1
n2–1
s2
2
n2
æ
è
çö
ø
÷
2=
2302
30 +1952
40
æ
è
çö
ø
÷
2
1
29
2302
30
æ
è
çö
ø
÷
2
+1
39
1952
40
æ
è
çö
ø
÷
2=56.5
Use df = 56
Using t table, p-value is between .005 and .01.
Exact p-value corresponding to t = 2.40 is .0099
2
5844 487
12
i
x
xn
= = =
12
xx−
= 525 – 487 = 38 points higher if parents are college grads
2
1
()52962 3530.8 59.42
i
xx
−
2
2
2
2
()29456 2677.82 51.75
1 12 1
i
xx
sn
−
= = = =
−−
( )
1 2 0
2 2 2 2
12
12
(525 487) 0 1.80
59.42 51.75
16 12
x x D
tss
nn
−− −−
= = =
+
+
22
22 22
12
12
2 2 2 2
2 2 2 2
12
1 1 2 2
59.42 51.75
16 12 25.3
1 59.42 1 51.75
11
15 16 11 12
11
ss
nn
df ss
n n n n
++
= = =
+
+
−−